Parametric Equations

(-5, 13)

(13, -5)

(5, 13)

(13, 5)

$$\frac{x^2}{9}+\frac{y^2}{16}=1$$

$$\frac{y^2}{9}+\frac{x^2}{16}=1$$

$$\cos^2\theta+\sin^2\theta=1$$

$$\frac{x^2}{9}-\frac{y^2}{16}=1$$

$$x=t^2,\ y=t-1$$  

$$x=t,\ y=t^2-1$$  

$$x=\sqrt{t},\ y=t-1$$  

$$x=\sqrt{t},\ y=t^2-1$$  

$$x=\cos t,\ y=\sin t$$  

$$x=-\cos t,\ y=\sin t$$  

$$x=\cos t,\ y=-\sin t$$  

$$x=-\cos t,\ y=-\sin t$$  

$$x=t,\ \ y=\frac{7}{2}\left(t-3\right)+2$$  where $$\ 1\le t\le3$$  

$$x=t,\ \ y=\frac{7}{2}\left(t-3\right)+2$$  where $$\ 0\le t\le1$$  

$$x=3-2t,\ y=2-7t$$  where $$\ 0\le t\le1$$  

$$x=3-2t,\ y=2-7t$$ where   $$1\le t\le3$$  

$$x=\cos t,\ y=\sin t$$  

$$x=\sec t,\ y=\tan t$$  

$$x=\tan t,\ y=\sin t$$  

$$x=\csc t,\ y=\cot t$$  

Every rectangular equation can be expressed as a parametric equation.

Every parametric equation can be expressed as a rectangular equation.

The best approach to eliminate the parameter is always to solve for the parameter from one equation and substitute to the other one.

If two parametrically defined curves intersects, they do so simultaneously.

The inverse function of $$x=t,\ y=-2t+1$$  is $$x=-2t+1,y=t$$ .

$$x=\cos\theta,\ y=\sin\theta$$  where $$0\le\theta\le\frac{3\pi}{2}$$  

$$x=\cos\theta,\ y=\sin\theta$$   where $$0\le\theta\le\pi$$

$$x=\cos\theta,\ y=\sin\theta$$   where $$\frac{\pi}{2}\le\theta\le\frac{3\pi}{2}$$

$$x=\cos\theta,\ y=\sin\theta$$   where $$\frac{\pi}{2}\le\theta\le2\pi$$